91Ó°ÊÓ

Consider the formulas:

$\begin{array}{r}\mathbf{Y}\mathbf{=}\frac{\mathbf{\text{Tensile stress}}}{\mathbf{\text{Tensile strain}}}\\ \mathbf{=}\frac{{\mathbf{F}}_{\mathbf{âŠ\yen }}\mathbf{/}\mathbf{A}}{\mathbf{Δ}\mathbf{I}\mathbf{/}{\mathbf{I}}_{\mathbf{0}}}\\ \mathbf{=}\frac{{\mathbf{F}}_{\mathbf{âŠ\yen }}}{\mathbf{Δ}\mathbf{V}}\frac{{\mathbf{I}}_{\mathbf{0}}}{\mathbf{A}}\end{array}$

Length of wire is $l_0$

Cross sectional area of object is A

Weight of object is W

(a)

Here we have to prove:$k=\frac{YA}{I_0}$

Our goal is to express$\frac{F_{âŠ\yen }}{∆l}$because that is the force constant which we can see in formula (11.7)

Now, by formula (11.7) which is given by,

$\mathrm{Hooke}\text{'}\mathrm{s}\mathrm{law}=\frac{\mathrm{Stress}}{\mathrm{Strain}}$

Here, stress is measure of forces applied to body. Strain is measure of how much deformation results from stress.

Now, from equation (1), we have,

$\begin{array}{c}Y=\frac{F_{âŠ\yen }}{A}\frac{I_0}{\mathrm{Δ}I}\\ \frac{YA}{I_0}=\frac{F_{âŠ\yen }}{\mathrm{Δ}I}\\ \mathrm{Now},\mathrm{we}\mathrm{know}\mathrm{that}\frac{{\mathrm{F}}_{âŠ\yen }}{∆\mathrm{l}}=\mathrm{k}\mathrm{So},\mathrm{above}\mathrm{equation}\mathrm{becomes}\\ k=\frac{YA}{I_0}\end{array}$

Hence proved.

(b)

From part (a) we have,

$k=\frac{YA}{I_0}$

Now, for copper wire above equation becomes

$k=\frac{Y_{Cu}A}{I_0}$ (2)

Now, $Y_{Cu}=11×{10}^{10}$

Determine the area:

$\begin{array}{r}A=\mathrm{π}{r}^2\\ =\mathrm{π}\left(\frac{1.29×{10}^{−3}}{2}\right)\\ =1.31×{10}^{−6}{\mathrm{m}}^2\end{array}$

Now, by substituting all the numerical values in equation (2).

$\begin{array}{r}k=\frac{\left(11×{10}^{10}\right)\left(1.31×{10}^{−6}{\mathrm{m}}^2\right)}{0.75\mathrm{m}}\\ =1.92×{10}^5\frac{\mathrm{N}}{\mathrm{m}}\end{array}$

Hence, the force constant is$1.92×{10}^5\frac{\mathrm{N}}{\mathrm{m}}$.

(c)

To find the weight W for given$∆l$is given by,

$\begin{array}{r}k=\frac{W}{∆l}\\ W=k\mathrm{Δ}l\\ W=\left(1.92×{10}^5\mathrm{N}/\mathrm{m}\right)â‹\dots \left(1.25×{10}^{−3}\mathrm{m}\right)\\ W=2.4×{10}^2\end{array}$

Hence, Weight is$2.4×{10}^2\mathrm{N}$ .